Free math worksheets

- What are all the individual results that could occur on one spin? - An event can contain one or more outcomes from the sample space.

1. The possible individual outcomes are R, G, and B, so \(S=\{R,G,B\}\). 2. An outcome is one possible result. The set \(\{G,B\}\) contains two outcomes, so it is an event.

a) \(S=\{R,G,B\}\). b) \(\{G,B\}\) is an event because it contains two possible outcomes.

- Read every sector label, then ask which distinct results can actually be recorded. - A set does not list the same outcome more than once.

1. Reading the spinner, the possible displayed numbers are \(1\), \(2\), \(3\), and \(4\). 2. A sample space lists distinct possible outcomes, not one entry for every physical sector. Therefore, \(S=\{1,2,3,4\}\).

\(S=\{1,2,3,4\}\). Repeated labels represent the same recorded outcome, so each possible number appears only once in the sample space.

- Pair each possible coin result with every possible number. - Because the two stages record different kinds of results, keep their order fixed.

1. The coin result can be H or T, and each can be paired with any of the three numbers. 2. The six ordered outcomes are \((H,1)\), \((H,2)\), \((H,3)\), \((T,1)\), \((T,2)\), and \((T,3)\).

\(S=\{(H,1),(H,2),(H,3),(T,1),(T,2),(T,3)\}\).

- List the favorable outcomes for each event before adding probabilities. - Remember that \(1\) is not prime. - Identify the perfect squares from \(1\) through \(8\). - For an “or” event, include every outcome satisfying at least one condition, without double-counting overlaps.

1. Event \(A\) contains \(1\) and \(4\), so \(P(A)=0.05+0.20=0.25\). 2. Event \(B\) contains \(5\), \(6\), \(7\), and \(8\), so \(P(B)=0.08+0.12+0.15+0.15=0.50\). 3. Event \(C\) contains the primes \(2\), \(3\), \(5\), and \(7\), so \(P(C)=0.15+0.10+0.08+0.15=0.48\). 4. The odd outcomes are \(1\), \(3\), \(5\), and \(7\), and the outcomes divisible by \(4\) are \(4\) and \(8\). These sets are disjoint, so \(P(D)=0.05+0.10+0.08+0.15+0.20+0.15=0.73\).

\(A\): \(P(A)=0.25\) \(B\): \(P(B)=0.50\) \(C\): \(P(C)=0.48\) \(D\): \(P(D)=0.73\)

- Recall the possible results of one die roll. - Translate “absolute difference” into an equation involving the two coordinates. - Check each possible first-roll value and determine which second-roll values make the difference \(2\). - Remember that order matters: \((1, 3)\) and \((3, 1)\) are different outcomes.

1. An outcome is an ordered pair \((a, b)\), where \(a, b \in \{1, 2, 3, 4, 5, 6\}\). 2. The condition \(X=2\) means \(|a-b|=2\). 3. Checking the possible first-roll values gives the pairs \((1, 3)\), \((2, 4)\), \((3, 1)\), \((3, 5)\), \((4, 2)\), \((4, 6)\), \((5, 3)\), and \((6, 4)\). 4. Therefore, \(E=\{(1, 3), (2, 4), (3, 1), (3, 5), (4, 2), (4, 6), (5, 3), (6, 4)\}\).

\(E=\{(1, 3), (2, 4), (3, 1), (3, 5), (4, 2), (4, 6), (5, 3), (6, 4)\}\)

- How many outcomes are possible when a coin is tossed three times? - Determine how many heads and tails make the difference “heads minus tails” equal to \(-1\). - Decide whether more than one arrangement has those numbers of heads and tails. - List every possible order.

1. There are \(2^3=8\) possible ordered triples of \(H\) and \(T\). 2. Let \(n_H\) be the number of heads and \(n_T\) the number of tails. Then \(n_H+n_T=3\) and \(n_H-n_T=-1\). 3. Substituting \(n_T=3-n_H\) gives \(n_H-(3-n_H)=-1\), so \(2n_H-3=-1\) and \(n_H=1\). Thus, the event contains outcomes with exactly one head and two tails. 4. The possible orders are \((H, T, T)\), \((T, H, T)\), and \((T, T, H)\). 5. Therefore, \(E=\{(H, T, T), (T, H, T), (T, T, H)\}\).

\(E=\{(H, T, T), (T, H, T), (T, T, H)\}\)

- What two conditions must every finite probability model satisfy? - Check the total probability before changing any value. - Only one probability is allowed to change.

1. Every listed probability is between \(0\) and \(1\), but their sum is \(0.25+0.35+0.20+0.25=1.05\), so the model is not valid. 2. The first three probabilities sum to \(0.80\). To make the total \(1\), \(P(D)\) must be \(1-0.80=0.20\).

No. The probabilities sum to \(1.05\). The model becomes valid if \(P(D)=0.20\).

- Ask whether heads followed by tails is the same ordered result as tails followed by heads. - After completing the sample space, select only outcomes with one H and one T.

1. The order of the two tosses matters. The outcome TH is possible but is missing, so the complete sample space is \(\{HH,HT,TH,TT\}\). 2. Exactly one head occurs in HT and TH, so \(E=\{HT,TH\}\).

a) The missing outcome is \(TH\), so \(S=\{HH,HT,TH,TT\}\). b) \(E=\{HT,TH\}\).

- What must the probabilities of all outcomes add to? - Write the relationship between the two unknown probabilities using one variable. - To find the probability of an event, add the probabilities of its outcomes.

1. Let \(P(x_4)=p\). Then \(P(x_3)=3p\). 2. Use the fact that all outcome probabilities add to \(1\): \(0.12+0.28+3p+p=1\). 3. Solve: \(0.40+4p=1\), so \(4p=0.60\) and \(p=0.15\). Therefore, \(P(x_4)=0.15\) and \(P(x_3)=0.45\). 4. Add the probabilities of the outcomes in \(E\): \(P(E)=0.12+0.45=0.57\).

\(P(x_3)=0.45\), \(P(x_4)=0.15\), and \(P(E)=0.57\).

- The probabilities of all outcomes must add to \(1\). - Which outcome is not in event \(A\)? Use a complement. - A set with three elements has \(2^3\) subsets. - An event is a subset of the sample space. Add the probabilities of the outcomes it contains.

1. Since \(A^c=\{\omega_3\}\), \(P(\omega_3)=1-P(A)=1-0.55=0.45\). 2. Since \(B^c=\{\omega_1\}\), \(P(\omega_1)=1-P(B)=1-0.75=0.25\). 3. The outcome probabilities add to \(1\), so \(P(\omega_2)=1-0.25-0.45=0.30\). 4. Add the appropriate outcome probabilities for each subset of \(S\): \(P(\emptyset)=0\), \(P(\{\omega_1\})=0.25\), \(P(\{\omega_2\})=0.30\), \(P(\{\omega_3\})=0.45\), \(P(\{\omega_1,\omega_2\})=0.55\), \(P(\{\omega_1,\omega_3\})=0.70\), \(P(\{\omega_2,\omega_3\})=0.75\), and \(P(S)=1\).

Individual outcomes: \(P(\omega_1)=0.25\), \(P(\omega_2)=0.30\), \(P(\omega_3)=0.45\) Events: \(P(\emptyset)=0\), \(P(\{\omega_1\})=0.25\), \(P(\{\omega_2\})=0.30\), \(P(\{\omega_3\})=0.45\), \(P(\{\omega_1,\omega_2\})=0.55\), \(P(\{\omega_1,\omega_3\})=0.70\), \(P(\{\omega_2,\omega_3\})=0.75\), \(P(S)=1\)

- Represent the ordered outcomes in a grid or organized list. - For part b), count every outcome with a \(7\) on the first die, then count the outcomes with a \(7\) on the second die that have not already been counted. - Identify the powers of \(2\) available on each die.

1. There are \(8\cdot12=96\) equally likely ordered outcomes. 2. A sum of \(10\) occurs for \((1, 9),(2, 8),(3, 7),(4, 6),(5, 5),(6, 4),(7, 3),(8, 2)\), so the probability is \(\frac{8}{96}=\frac{1}{12}\). 3. For at least one \(7\), there are \(12\) outcomes with the eight-sided die showing \(7\). There are \(7\) additional outcomes with the twelve-sided die showing \(7\) and the eight-sided die showing a value other than \(7\). Thus there are \(12+7=19\) favorable ordered outcomes, so the probability is \(\frac{19}{96}\). 4. A product is a power of \(2\) only when each factor is a power of \(2\). Each die can show \(1,2,4,8\), giving \(4\cdot4=16\) favorable outcomes. Thus, the probability is \(\frac{16}{96}=\frac{1}{6}\).

a) \(\frac{1}{12}\) b) \(\frac{19}{96}\) c) \(\frac{1}{6}\)

- Count ordered triples for the small possible sums. - Use the complement of no \(8\). - For all different values, count the remaining choices at each roll.

1. There are \(8^3=512\) equally likely ordered outcomes. Sums less than \(6\) are \(3\), \(4\), and \(5\), occurring in \(1\), \(3\), and \(6\) ordered outcomes, respectively. Thus, \(P(S<6)=\frac{10}{512}=\frac{5}{256}\). 2. The probability of no \(8\) in three rolls is \(\left(\frac{7}{8}\right)^3\). Therefore, \(P(\text{at least one }8)=1-\frac{343}{512}=\frac{169}{512}\). 3. The first result can be any of \(8\) values, the second any of the remaining \(7\), and the third any of the remaining \(6\). Thus, \(P(\text{all different})=\frac{8\cdot7\cdot6}{8^3}=\frac{21}{32}\).

1. \(\frac{5}{256}\approx 0.01953\) 2. \(\frac{169}{512}\approx 0.33008\) 3. \(\frac{21}{32}=0.65625\)

- Treat each ordered pair of die results as one equally likely outcome. - For the prime-sum event, focus on which ordered outcomes make an allowable sum. - For the claim about sums, compare how many ordered outcomes can produce different sum values.

1. The \(36\) ordered pairs are equally likely. 2. Prime sums are \(2,3,5,7,11\). Counting their ordered pairs gives \(1+2+4+6+2=15\), so \(P(E_1)=\frac{15}{36}=\frac{5}{12}\). 3. Exactly \(15\) ordered pairs have a product divisible by \(4\), so \(P(E_2)=\frac{15}{36}=\frac{5}{12}\). 4. Jisoo's claim is false. The ordered pairs are equally likely, but the sums are not. For example, a sum of \(2\) has one ordered pair, while a sum of \(7\) has six.

1. \(P(E_1)=\frac{5}{12}\) and \(P(E_2)=\frac{5}{12}\) 2. The claim is false because different sums occur in different numbers of equally likely ordered pairs.

- Use ordered pairs to represent the \(64\) equally likely outcomes. - For the absolute-difference event, count pairs systematically from the smaller result to the larger result. - For part 2, describe the complement in terms of the largest value either die may show.

1. There are \(64\) equally likely ordered pairs. An absolute difference of \(2\) occurs in \(12\) pairs, so the probability is \(\frac{12}{64}=\frac{3}{16}=0.1875\). 2. For the second event, its complement is that both results are from \(1\) through \(6\). That complement contains \(6\cdot6=36\) ordered pairs, so the required probability is \(1-\frac{36}{64}=\frac{7}{16}=0.4375\).

1. \(\frac{3}{16}=0.1875\) 2. \(\frac{7}{16}=0.4375\)

- Count the total ordered outcomes for two die rolls. - Organize the possible sums in a table or grid. - Translate “at most,” “at least,” and “less than” carefully. - Identify the prime numbers from \(2\) through \(12\). - Count the grid entries that satisfy each event.

1. There are \(6\cdot6=36\) equally likely ordered outcomes. 2. For part a), the event is \(4\le X<7\). Sums \(4\), \(5\), and \(6\) occur in \(3\), \(4\), and \(5\) ways, respectively, so \(P(4\le X<7)=\frac{12}{36}=\frac{1}{3}\). 3. For part b), the prime sums are \(2\), \(3\), \(5\), \(7\), and \(11\). They occur in \(1+2+4+6+2=15\) ways, so \(P(X\in\{2,3,5,7,11\})=\frac{15}{36}=\frac{5}{12}\). 4. For part c), the event is \(X\le3\lor X>10\). Sums \(2\), \(3\), \(11\), and \(12\) occur in \(1+2+2+1=6\) ways, so the probability is \(\frac{6}{36}=\frac{1}{6}\).

a) \(P(4\le X<7)=\frac{1}{3}\) b) \(P(X\in\{2,3,5,7,11\})=\frac{5}{12}\) c) \(P(X\le3\lor X>10)=\frac{1}{6}\)

- Interpret the interval \([5, 8]\) as a set of possible integer values of \(S\). - For part b), identify sums satisfying both conditions. - For part c), consider counting the excluded sums and subtracting from \(1\). - Recall the symmetric pattern of counts for sums of two fair dice.

1. There are \(36\) equally likely ordered outcomes. 2. For part a), the event is \(5\le S\le8\). Sums \(5\), \(6\), \(7\), and \(8\) occur in \(4+5+6+5=20\) ways, so \(P(5\le S\le8)=\frac{20}{36}=\frac{5}{9}\). 3. For part b), the odd sums greater than \(8\) are \(9\) and \(11\). They occur in \(4+2=6\) ways, so \(P(S>8\land S\text{ is odd})=\frac{6}{36}=\frac{1}{6}\). 4. For part c), use the complement of \(S\in\{7, 11\}\). These sums occur in \(6+2=8\) ways, so \(P(S\notin\{7, 11\})=1-\frac{8}{36}=\frac{7}{9}\).

a) \(P(5\le S\le8)=\frac{5}{9}\) b) \(P(S>8\land S\text{ is odd})=\frac{1}{6}\) c) \(P(S\notin\{7, 11\})=\frac{7}{9}\)

- Use the \(144\) ordered pairs as the sample space. - List favorable pairs systematically. - For square products, consider the prime-factor parity of each factor.

1. There are \(12\cdot12=144\) equally likely ordered pairs. 2. A sum of \(15\) occurs in \(10\) ordered pairs, so \(P=\frac{10}{144}=\frac{5}{72}\). 3. Exactly \(22\) ordered pairs have a perfect-square product, so \(P=\frac{22}{144}=\frac{11}{72}\). 4. An absolute difference of at least \(10\) occurs for \((1, 11),(1, 12),(2, 12)\) and their reverses, giving \(6\) outcomes. Thus, \(P=\frac{6}{144}=\frac{1}{24}\). 5. Prime sums from \(2\) through \(24\) are \(2,3,5,7,11,13,17,19,23\). They occur in \(1+2+4+6+10+12+8+6+2=51\) ordered pairs, so \(P=\frac{51}{144}=\frac{17}{48}\).

1. \(\frac{5}{72}\approx 0.06944\) 2. \(\frac{11}{72}\approx 0.15278\) 3. \(\frac{1}{24}\approx 0.04167\) 4. \(\frac{17}{48}\approx 0.35417\)